Question

Let $f(x)=\frac{5}{2}x^{2}-e^{x}$. Then the value of $c$ such that $f''(c)=0$ is

• A. 1
• B. $\log 5$
• C. $5e$
• D. $e^{5}$
• E. 0

Hint:

Algebra Tip: In advanced mathematics and calculus contexts, $\log(x)$ without a specified base almost always implies the natural logarithm ($\ln x$).

Answer 1

The Correct Option is B

Concept:
The notation $f^{\prime\prime}(c) = 0$ means we must find the second derivative of the function, set it equal to zero, and solve for the corresponding $x$-value (which is $c$). This point represents an inflection point on the graph.

Step 1: Find the first derivative $f^{\prime(x)$.}
Differentiate the given function $f(x) = \frac{5}{2}x^2 - e^x$ with respect to $x$: $$f^{\prime}(x) = \frac{5}{2}(2x) - e^x$$ $$f^{\prime}(x) = 5x - e^x$$

Step 2: Find the second derivative $f^{\prime\prime(x)$.}
Differentiate the first derivative again: $$f^{\prime\prime}(x) = 5(1) - e^x$$ $$f^{\prime\prime}(x) = 5 - e^x$$

Step 3: Set the second derivative to zero.
We are given the condition $f^{\prime\prime}(c) = 0$. Substitute $x = c$: $$5 - e^c = 0$$

Step 4: Isolate the exponential term.
Add $e^c$ to both sides of the equation: $$e^c = 5$$

Step 5: Solve for c using logarithms.
Take the natural logarithm ($\ln$, often denoted as $\log$ in higher mathematics) of both sides to cancel the base $e$: $$\ln(e^c) = \ln(5)$$ $$c = \log 5$$ Hence the correct answer is (B) $\log 5$.