Free Math Worksheets

- How does a horizontal shift change the input of a function, while a vertical shift changes its output? - How is a reflection across the x-axis represented in a function equation? - Apply the transformations in the stated order.

1. Reflecting across the x-axis gives \(-e^x\). 2. A vertical stretch by a factor of \(2.5\) gives \(-2.5e^x\). 3. Shifting \(3\) units left replaces \(x\) with \(x+3\), giving \(-2.5e^{x+3}\). 4. Shifting \(1\) unit down gives \(g(x)=-2.5e^{x+3}-1\).

\(g(x)=-2.5e^{x+3}-1\)

- A term containing no \(x\) has derivative \(0\). - An added constant produces a vertical translation. - An x-intercept has y-coordinate \(0\). - Recall that \(e^0=1\).

1. Because \(2t\) is constant with respect to \(x\), \(g_t'(x)=e^x\). 2. The graph is shifted vertically by \(2t\) units. It shifts up when \(t>0\) and down when \(t<0\). 3. Crossing the x-axis at \(x=0\) requires \(g_t(0)=0\). Thus, \(1+2t=0\), so \(t=-\frac{1}{2}\).

1. \(g_t'(x)=e^x\) 2. A vertical shift of \(2t\) units. 3. \(t=-\frac{1}{2}\)

- Express \(f(x+5)\) in terms of \(f(x)\) using exponent properties. - Compare the new function value with the original by forming a ratio. - Rewrite \(\frac{1}{9}\) as a power of \(3\). - Set up an equation with the same base on both sides.

1. a) Compare the new and original values: \(\frac{f(x+5)}{f(x)}=\frac{3^{x+5}}{3^x}=3^5=243\). The function value is multiplied by \(243\). 2. b) Let the input decrease by \(k\). Then \(\frac{f(x-k)}{f(x)}=\frac{3^{x-k}}{3^x}=3^{-k}\). 3. Set \(3^{-k}=\frac{1}{9}=3^{-2}\). Therefore, \(k=2\), so the input must decrease by \(2\).

a) A factor of \(243\) b) Decrease the input by \(2\).

- How do changes inside the input differ from constants added to the entire function? - Start with the horizontal asymptote of \(e^x\) and apply the vertical shift. - What is the x-coordinate of every point on the y-axis?

1. Shifting \(2\) units right replaces \(x\) with \(x-2\), and shifting \(5\) units down subtracts \(5\). Thus, \(g(x)=e^{x-2}-5\). 2. The horizontal asymptote of \(e^x\) is \(y=0\). Shifting the graph down \(5\) units changes the asymptote to \(y=-5\). 3. At the y-intercept, \(x=0\). Therefore, \(g(0)=e^{-2}-5\), so the y-intercept is \((0, e^{-2}-5)\).

a) \(g(x)=e^{x-2}-5\) b) \(y=-5\) c) \((0, e^{-2}-5)\)

- How does a vertical stretch affect the entire function expression? - What change to the input shifts a graph to the right? - What x-value identifies a y-intercept?

1. A vertical stretch by a factor of \(2\) gives \(2e^x\). 2. Shifting \(3\) units right replaces \(x\) with \(x-3\), so \(g(x)=2e^{x-3}\). 3. At the y-intercept, \(x=0\). Therefore, \(g(0)=2e^{-3}\), and the y-intercept is \((0, 2e^{-3})\).

a) \(g(x)=2e^{x-3}\) b) \((0, 2e^{-3})\)

- Evaluate each equation at \(x=0\) to compare y-intercepts. - Determine which graphs increase and which decrease. - Compare horizontal asymptotes and end behavior. - Identify reflections and vertical shifts.

1. Graph \(A\) is increasing, stays above the x-axis, has y-intercept \((0, 1)\), and approaches \(y=0\) as \(x\to-\infty\). Thus, \(A\) matches (1), \(y=e^x\). 2. Graph \(B\) is decreasing, stays above the x-axis, has y-intercept \((0, 1)\), and approaches \(y=0\) as \(x\to\infty\). Thus, \(B\) matches (2), \(y=e^{-x}\). 3. Graph \(C\) is the basic exponential graph shifted down \(3\) units. It has y-intercept \((0, -2)\) and horizontal asymptote \(y=-3\). Thus, \(C\) matches (3), \(y=e^x-3\). 4. Graph \(D\) is decreasing, has y-intercept \((0, 1)\), and approaches \(y=2\) as \(x\to-\infty\). Thus, \(D\) matches (4), \(y=-e^x+2\).

\(A\): (1) \(B\): (2) \(C\): (3) \(D\): (4)

- Use the band width to find the upper and lower boundaries. - Evaluate the function at each specified x-value. - Compare each function value with both band boundaries.

1. A total width of \(0.4\) means the band extends \(0.2\) above and below \(y=1\). Its boundaries are \(y=0.8\) and \(y=1.2\). 2. \(f(3)=1+2(0.5)^3=1.25\). Since \(1.25>1.2\), the graph is outside the band at \(x=3\). 3. \(f(4)=1+2(0.5)^4=1.125\). Since \(0.8\leq1.125\leq1.2\), the graph is inside the band at \(x=4\).

At \(x=3\), the graph is outside the band because \(f(3)=1.25\). At \(x=4\), it is inside the band because \(f(4)=1.125\).

- How can you find where a graph intersects an axis? - What happens to \(e^x\) as \(x\) becomes very negative? - How does reflection across the x-axis affect whether a graph increases or decreases? - Does a reflection and vertical shift create a change in concavity for an exponential graph? - Recall the value of \(e^0\).

1. Set \(f(x)=0\): \(5-e^x=0\), so \(e^x=5\) and \(x=\ln(5)\). Statement a is true. 2. As \(x\to-\infty\), \(e^x\to0\), so \(5-e^x\to5\). Statement b is true. 3. The graph of \(e^x\) is strictly increasing. Reflecting it across the x-axis makes \(-e^x\) strictly decreasing, and shifting it upward does not change monotonicity. Statement c is false. 4. The graph of \(-e^x\) is concave down everywhere, and a vertical shift does not change concavity. Therefore, there is no inflection point. Statement d is false. 5. Since \(f(0)=5-e^0=5-1=4\), the y-intercept is \((0, 4)\). Statement e is true.

a) True; solving \(5-e^x=0\) gives \(x=\ln(5)\). b) True; \(f(x)\to5\) as \(x\to-\infty\). c) False; \(f\) is strictly decreasing. d) False; the graph is concave down everywhere. e) True; \(f(0)=4\).

- First factor the coefficient of \(x\) from the entire exponent. - How does a coefficient multiplying \(x\) inside the exponent affect horizontal scaling? - What does the negative coefficient outside the exponential function do to the graph? - Which constant determines the horizontal asymptote?

1. Rewrite the exponent by factoring: \(0.5x-1=0.5(x-2)\), so \(g(x)=-2e^{0.5(x-2)}+4\). 2. Replacing \(x\) with \(0.5x\) stretches the graph horizontally by a factor of \(2\). 3. Replacing \(x\) with \(x-2\) after that stretch shifts the graph \(2\) units right, producing \(e^{0.5(x-2)}\). 4. Multiplying by \(-2\) stretches the graph vertically by a factor of \(2\) and reflects it across the x-axis. 5. Adding \(4\) shifts the graph \(4\) units up. The result is \(g(x)=-2e^{0.5(x-2)}+4\).

1. Stretch horizontally by a factor of \(2\). 2. Shift \(2\) units right. 3. Stretch vertically by a factor of \(2\). 4. Reflect across the x-axis. 5. Shift \(4\) units up.

- Rewrite the exponent so the horizontal shift is visible. - What transformation is represented by a negative sign on \(x\) inside the exponent? - How does the constant added outside the exponential expression affect the horizontal asymptote? - What does multiplying the entire exponential expression by \(-1\) do?

1. Rewrite the exponent as \(3-x=-(x-3)\). 2. Replacing \(x\) with \(-x\) reflects the graph of \(e^x\) across the y-axis, producing \(e^{-x}\). 3. Replacing \(x\) with \(x-3\) shifts that graph \(3\) units right, producing \(e^{-(x-3)}=e^{3-x}\). 4. Multiplying by \(-1\) reflects the graph across the x-axis, producing \(-e^{3-x}\). 5. Adding \(5\) shifts the graph \(5\) units up, producing \(k(x)=-e^{3-x}+5\). 6. As \(x\to\infty\), \(e^{3-x}\to0\), so \(k(x)\to5\). The horizontal asymptote is \(y=5\).

One possible sequence is: reflect across the y-axis, shift \(3\) units right, reflect across the x-axis, and shift \(5\) units up. The horizontal asymptote is \(y=5\).

- Test whether \(g(-x)=g(x)\). - First determine where the exponent \(-x^2+1\) is greatest. - Analyze the exponent as \(|x|\) becomes large.

1. \(g(-x)=2^{-(-x)^2+1}=2^{-x^2+1}=g(x)\), so \(g\) is even and its graph is symmetric about the y-axis. 2. Because the base \(2\) is greater than \(1\), \(g(x)\) is largest when the exponent \(-x^2+1\) is largest. This occurs at \(x=0\), where \(g(0)=2\). The highest point is \((0, 2)\). 3. As \(x\to\infty\) or \(x\to-\infty\), the exponent \(-x^2+1\to-\infty\). Therefore, \(g(x)\to0\) in both directions. 4. Geometrically, the x-axis, \(y=0\), is a horizontal asymptote on both ends of the graph.

a) Yes; \(g(-x)=g(x)\), so the graph is symmetric about the y-axis. b) \((0, 2)\) c) \(g(x)\to0\) as \(x\to\infty\) and as \(x\to-\infty\); \(y=0\) is a horizontal asymptote.

- How does a horizontal shift change the input of the function? - Use the product rule for powers with the same base. - Rewrite the constant scale factor as a power of \(e\). - How does the sign in \(x-c\) determine the direction of a horizontal shift?

1. A shift \(3\) units left replaces \(x\) with \(x+3\), so \(g(x)=e^{x+3}\). 2. Using exponent rules, \(e^{x+3}=e^3e^x=e^3f(x)\). Therefore, the same graph is produced by a vertical stretch with scale factor \(k=e^3\). 3. For part b, \(h(x)=\frac{1}{e^2}e^x=e^{-2}e^x=e^{x-2}=f(x-2)\). 4. Replacing \(x\) with \(x-2\) shifts the graph \(2\) units right.

a) \(g(x)=e^3e^x\), so \(k=e^3\). b) Shift \(G_f\) \(2\) units to the right.

- Use a transformed exponential function of the form \(ae^x+c\). - The vertical shift determines the horizontal asymptote. - Substitute \(x=0\) to use the y-intercept. - Since \(e^0=1\), the intercept gives a simple equation for \(a\).

1. Use the form \(F(x)=ae^x+c\), whose horizontal asymptote is \(y=c\). 2. For part a, \(c=5\). The y-intercept condition gives \(a+5=6\), so \(a=1\). One solution is \(F(x)=e^x+5\). 3. For part b, \(c=-2\). The y-intercept condition gives \(a-2=3\), so \(a=5\). One solution is \(F(x)=5e^x-2\).

a) \(F(x)=e^x+5\) b) \(F(x)=5e^x-2\)

- Recall the end behavior of \(e^x\). - Consider the effect of multiplying by a negative number. - The vertical shift determines the horizontal asymptote. - Set \(x=0\) to find the y-intercept.

1. As \(x\to\infty\), \(e^x\to\infty\), so \(f(x)\to-\infty\). As \(x\to-\infty\), \(e^x\to0\), so \(f(x)\to4\). Thus, \(y=4\) is a horizontal asymptote. 2. Since \(e^x\) is strictly increasing, \(-2e^x\) is strictly decreasing. Adding \(4\) does not change monotonicity, so \(f\) is strictly decreasing. The function approaches but never reaches \(4\), and it is unbounded below, so its range is \((-\infty, 4)\). 3. \(f(0)=4-2e^0=2\), so the y-intercept is \((0, 2)\).

1. \(\lim_{x\to\infty}f(x)=-\infty\); \(\lim_{x\to-\infty}f(x)=4\) 2. Strictly decreasing; range \((-\infty, 4)\) 3. \((0, 2)\)

- Factor the coefficient of \(x\) from the entire exponent. - A coefficient multiplying \(x\) inside the exponent changes horizontal scale. - Factors and constants outside the exponential affect outputs. - Check the sequence by reconstructing the final exponent.

1. Factor the exponent: \(0.5x-2=0.5(x-4)\). 2. Horizontally stretch the graph of \(g\) by a factor of \(2\), producing \(e^{0.5x}\). 3. Shift right \(4\) units, producing \(e^{0.5(x-4)}\). 4. Vertically stretch by a factor of \(4\). 5. Shift down \(1\) unit.

Horizontally stretch by a factor of \(2\), shift right \(4\) units, vertically stretch by a factor of \(4\), and shift down \(1\) unit.

- Set up an equation using \(h(x+k)=100h(x)\). - Solve an equation of the form \(a^k=b\) with logarithms. - Express \(36\) as a power of \(6\). - Use the power-of-a-power property to compare the two equations.

1. a) Set \(h(x+k)=100h(x)\): \(6^{x+k}=100\cdot6^x\). 2. Divide by \(6^x\): \(6^k=100\). Therefore, \(k=\log_6(100)=\frac{\log(100)}{\log(6)}\approx2.570\). 3. b) For \(g\), the equation is \(36^k=100\). Since \(36=6^2\), \(6^{2k}=100\). 4. Compared with \(6^{k_{\!6}}=100\), this gives \(2k_{\!36}=k_{\!6}\). Thus, \(k_{\!36}=\frac{1}{2}k_{\!6}\approx1.285\).

a) \(k\approx2.570\) b) The required increase is halved: \(k\approx1.285\).

- How can a positive number be written using \(e\) and the natural logarithm? - What change to the input reflects a graph across the y-axis? - Recall the meaning of a negative exponent.

1. Since \(4=e^{\ln(4)}\), \(f(x)=\left(e^{\ln(4)}\right)^x=e^{x\ln(4)}\). 2. Reflection across the y-axis replaces \(x\) with \(-x\), so \(g(x)=f(-x)=4^{-x}\). 3. Using exponent rules, \(4^{-x}=\left(4^{-1}\right)^x=\left(\frac{1}{4}\right)^x\). 4. In base \(e\), \(g(x)=e^{-x\ln(4)}\).

1. \(f(x)=e^{x\ln(4)}\) 2. \(g(x)=\left(\frac{1}{4}\right)^x=e^{-x\ln(4)}\)

- Which parameter determines the horizontal asymptote? - Find where \(y=x+1\) crosses the y-axis. - Substitute the known point into the general equation. - Recall the value of \(e^0\).

1. For \(h(x)=ae^x+c\), the horizontal asymptote is \(y=c\). Therefore, \(c=-2\). 2. The line \(y=x+1\) crosses the y-axis at \((0, 1)\), so \(h(0)=1\). 3. Substitute into \(h(x)=ae^x-2\): \(1=ae^0-2=a-2\). 4. Thus, \(a=3\), and \(h(x)=3e^x-2\).

\(a=3\), \(c=-2\), and \(h(x)=3e^x-2\)

- Write the scale factor \(3\) as a power with base \(e\). - Combine powers with the same base. - What direction is represented by replacing \(x\) with \(x+c\)?

1. Write \(3=e^{\ln(3)}\). Then \(h(x)=e^{\ln(3)}e^x=e^{x+\ln(3)}\). 2. Since \(h(x)=f(x+\ln(3))\), the graph of \(f\) is shifted \(\ln(3)\) units to the left.

Shift the graph \(\ln(3)\) units to the left.

- How does reflection across the x-axis change each output? - Apply the second transformation to the entire result of the first. - Compare the final expressions for all x-values.

1. Reflecting \(f\) across the x-axis gives \(-e^x\). Shifting that graph \(5\) units down gives \(h_1(x)=-e^x-5\). 2. Reversing the order, first shift \(f\) down \(5\) units to obtain \(e^x-5\). 3. Reflecting the entire shifted function across the x-axis gives \(h_2(x)=-(e^x-5)=-e^x+5\). 4. Since \(-e^x-5\ne-e^x+5\), the two orders produce different functions.

a) \(h_1(x)=-e^x-5\) b) \(h_2(x)=-e^x+5\). The order matters because \(h_1\ne h_2\).

- Apply each transformation to the current function, not just to the original exponential term. - Use parentheses when a vertical stretch follows a vertical shift. - Recall the end behavior of \(e^x\) as \(x\to-\infty\).

1. For the first order, shift and then scale: \(g_1(x)=2(e^x+3)=2e^x+6\). 2. Since \(e^x\to0\) as \(x\to-\infty\), the horizontal asymptote of \(g_1\) is \(y=6\). Also, \(g_1(0)=2e^0+6=8\), so the y-intercept is \((0, 8)\). 3. For the second order, scale and then shift: \(g_2(x)=2e^x+3\). 4. The horizontal asymptote of \(g_2\) is \(y=3\). Also, \(g_2(0)=2e^0+3=5\), so the y-intercept is \((0, 5)\).

1) \(g_1(x)=2e^x+6\); horizontal asymptote: \(y=6\); y-intercept: \((0, 8)\) 2) \(g_2(x)=2e^x+3\); horizontal asymptote: \(y=3\); y-intercept: \((0, 5)\)

- A reflection across the x-axis multiplies the entire current output by \(-1\). - Pay attention to whether the vertical shift is inside or outside the reflection. - Use the fact that \(e^x\) is always positive.

1. Reflect first and then shift: \(g(x)=-e^x+2\). 2. Solve \(-e^x+2=0\): \(e^x=2\), so \(x=\ln(2)\). 3. Shift first and then reflect: \(h(x)=-(e^x+2)=-e^x-2\). 4. Since \(e^x>0\) for every real \(x\), \(-e^x-2<0\). Therefore, \(h\) has no real zeros.

\(g(x)=-e^x+2\) has the zero \(x=\ln(2)\). \(h(x)=-e^x-2\) has no real zeros.

- Evaluate the function at \(x=0\). - Read each y-intercept from the graph. - Differentiate while treating \(a\) as a constant.

1. The y-intercept is \(f_a(0)=a-2\), so \(a\) equals the y-intercept plus \(2\). 2. Graph p has y-intercept \(1\), so \(a=3\). Graph q has y-intercept \(0\), so \(a=2\). Graph r has y-intercept \(-1\), so \(a=1\). Graph s has y-intercept \(-1.5\), so \(a=0.5\). 3. Increasing \(a\) moves the y-intercept upward. Since \(f_a'(x)=ae^x\), increasing \(a\) also increases the slope at every fixed \(x\), making the graph steeper.

a) p: \(a=3\); q: \(a=2\); r: \(a=1\); s: \(a=0.5\) b) The y-intercept moves upward, and the graph becomes steeper. c) \(f_a'(x)=ae^x\)

- Compare the horizontal asymptotes of the three graphs. - Track the point \((0, 1)\) on the parent function. - Determine which transformed graph is decreasing. - Use the form \(ae^{x-d}+c\) to organize the transformations.

1. The graph of \(g\) has horizontal asymptote \(y=-2\), so \(f\) is shifted down \(2\) units. The point \((0, 1)\) on \(f\) corresponds to \((1, -1)\) on \(g\), showing a shift right \(1\) unit. 2. Therefore, \(g(x)=e^{x-1}-2\). 3. The graph of \(h\) is decreasing and has horizontal asymptote \(y=3\). Reflecting \(f\) across the x-axis gives \(-e^x\), and shifting up \(3\) units gives the displayed graph. 4. Therefore, \(h(x)=-e^x+3\).

\(g\): shift right \(1\) unit and down \(2\) units; \(g(x)=e^{x-1}-2\) \(h\): reflect across the x-axis and shift up \(3\) units; \(h(x)=-e^x+3\)

- Compare where the two graphs level off to identify the vertical translation. - Track the point \((0, 1)\) from the parent graph. - A shift left places a plus sign with the horizontal shift inside the exponent.

1. The horizontal asymptote moves from \(y=0\) to \(y=-2\), so the graph shifts down \(2\) units. 2. The point \((0, 1)\) on \(f\) moves to \((-3, -1)\) on \(g\), so the graph shifts left \(3\) units. 3. Thus, \(g(x)=f(x+3)-2=e^{x+3}-2\). 4. Evaluate at \(x=0\): \(g(0)=e^3-2\approx18.09\).

1) Shift left \(3\) units and down \(2\) units. 2) \(g(x)=e^{x+3}-2\) 3) \(g(0)=e^3-2\approx18.09\)

- Evaluate the inner function first. - Work backward from the output of the outer function when solving an equation. - Changes inside the exponent produce horizontal shifts; changes outside produce vertical shifts.

1. Since \(v(3)=1\) and \(u(1)=2\), \(f(3)=2\). Since \(u(1)=2\) and \(v(2)=0\), \(g(1)=0\). 2. The equation \(f(x)=1\) means \(u(v(x))=1\). Since \(u(0)=1\), solve \(v(x)=0\), which gives \(x=2\). The equation \(g(x)=0\) means \(v(u(x))=0\). Since \(v(2)=0\), solve \(u(x)=2\), which gives \(x=1\). 3. Since \(f(x)=2^{x-2}\), its graph is the graph of \(u\) shifted \(2\) units right. Since \(g(x)=2^x-2\), its graph is the graph of \(u\) shifted \(2\) units down.

a) \(f(3)=2\) and \(g(1)=0\) b) \(f(x)=1\) at \(x=2\); \(g(x)=0\) at \(x=1\) c) \(f\): shift \(2\) units right; \(g\): shift \(2\) units down

- Write an equation for each composition. - Identify reflections and horizontal or vertical shifts. - Compare y-intercepts and end behavior.

1. \(f(g(x))=2^{-x}\) is a decreasing exponential function with y-intercept \(1\), so it is graph p. 2. \(g(f(x))=-2^x\) is the graph of \(2^x\) reflected across the x-axis, so it is graph q. 3. \(f(h(x))=2^{x-3}\) is the graph of \(2^x\) shifted \(3\) units right, so it is graph r. 4. \(h(f(x))=2^x-3\) is the graph of \(2^x\) shifted \(3\) units down, so it is graph s.

p) \(k_a(x)=f(g(x))\) q) \(k_b(x)=g(f(x))\) r) \(k_c(x)=f(h(x))\) s) \(k_d(x)=h(f(x))\)

- Substitute \(x=0\) to compare y-intercepts. - Identify each horizontal asymptote. - Compare growth rates at the same positive x-value. - Use exponent rules to identify equivalent equations.

1. Graph 1 has y-intercept \(2\) and grows like a vertical scaling of \(e^x\). It matches \(y=2e^x\). 2. Graph 2 also has y-intercept \(2\), but its horizontal asymptote is \(y=1\). It matches \(y=e^x+1\). 3. Graph 3 has y-intercept \(e\) and passes through \((-1, 1)\). It matches \(y=e^{x+1}\). Since \(e^{x+1}=e\cdot e^x\), both listed equations describe this graph. 4. Graph 4 has y-intercept \(-1\) and horizontal asymptote \(y=-2\). It matches \(y=e^x-2\).

Graph 1: \(y=2e^x\) Graph 2: \(y=e^x+1\) Graph 3: \(y=e^{x+1}\) and \(y=e\cdot e^x\) Graph 4: \(y=e^x-2\)

- How does horizontal scaling change the input of a function? - Which exponent rule rewrites \(e^{x/2}\) as a power raised to \(x\)? - Express \(5\) as a power with base \(e\). - Use the inverse relationship between exponential and logarithmic functions.

1. A horizontal stretch by a factor of \(d\) replaces \(x\) with \(\frac{x}{d}\). For \(d=2\), \(g(x)=f\!\left(\frac{x}{2}\right)=e^{x/2}\). 2. Rewrite \(e^{x/2}=\left(e^{1/2}\right)^x=(\sqrt{e})^x\). Thus, \(b=\sqrt{e}\). 3. To produce \(h(x)=5^x\), require \(e^{x/d}=5^x\). Since \(5^x=e^{x\ln(5)}\), the exponents must satisfy \(\frac{1}{d}=\ln(5)\). 4. Therefore, \(d=\frac{1}{\ln(5)}\approx0.621\). Because \(d<1\), this is a horizontal compression.

a) \(g(x)=(\sqrt{e})^x\), so \(b=\sqrt{e}\). b) \(d=\frac{1}{\ln(5)}\approx0.621\)

- The vertical shift determines the horizontal asymptote. - Choose an even polynomial exponent for y-axis symmetry. - Make the exponent approach \(-\infty\) at both ends. - Use the given point to determine the remaining coefficient.

1. Choose \(c=1\) so that \(y=1\) can be the horizontal asymptote. 2. Choose the even polynomial \(g(x)=-x^2\). Then \(g(x)\to-\infty\) as \(x\to\pm\infty\), so \(e^{g(x)}\to0\), and the resulting function is even. 3. Since \(g(0)=0\), the condition \(h(0)=-1\) gives \(a+1=-1\), so \(a=-2\). 4. One valid function is \(h(x)=-2e^{-x^2}+1\). It is even, approaches \(1\) as \(x\to\pm\infty\), and satisfies \(h(0)=-1\). 5. Its derivatives are \(h'(x)=4xe^{-x^2}\) and \(h''(x)=4(1-2x^2)e^{-x^2}\). Since \(h'(0)=0\) and \(h''(0)=4>0\), \((0,-1)\) is a local minimum.

One valid choice is \(a=-2\), \(c=1\), and \(g(x)=-x^2\), giving \(h(x)=-2e^{-x^2}+1\).