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How do I find concave up and concave down from $f(x) = {x^3} + 3{x^2} + 5x + 7$?

Last updated date: 13th Jun 2024
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Hint: Start by considering $f(x)$ as the function of $x$. Next step is evaluating the first and second derivative. After the second derivation, check for the signs that have changed. Then finally evaluate all the conditions and then decide if the graph is downwards or upwards.

Complete step by step solution:
First we will start off by evaluating the value of the second
f(x)\,\,\, = {x^3} + 3{x^2} + 5x + 7 \\
f'(x) = 3{x^2} + 3(2){x^1} + 5 + 0 \\
f'(x) = 3{x^2} + 6x + 5 \\
Now we will evaluate the value of the second derivative.
f'(x) = 3{x^2} + 6x + 5 \\
f''(x) = 3(2){x^1} + 6(1) + 0 \\
f''(x) = 6x + 6 \\
f''(x) = 6(x + 1) \\
Here, the second derivative changes sign from negative to positive as $x$ increases through the value at $x = 1$.
Hence, the graph of $f(x)$ is concave down when $x < 1$, and the graph of $f(x)$ is concave up when $x > 1$ and this graph has an inflection point at $x = 1$.

Additional Information: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.

Note: While applying the power rule make sure you have considered the power with their respective signs. Remember that the derivative of ${x^3}$ is $3{x^2}$, the derivative of ${x^2}$ is $2x$ and the derivative of a constant is zero. While applying the product rule, keep the first term as it is and differentiate the second term, then differentiate the first term and keep the second term as it is or vice versa.