Animated Solution for Mathematics - Differentiation: The curve $y(x)=a{x}^3+b{x}^2+cx+5$ touches the x-axis at the point $P(-2,0)$ and cuts the y-axis at the point $Q$, where $y^{\prime }$ is equal to 3. Then the local maximum value of $y(x)$ is :

Imagine you are standing on a coordinate plane, watching a cubic curve dance across the x-axis. It does not just cut through; it approaches, kisses the axis at $P(-2,0)$, and then gracefully turns away.

In the language of calculus, this tangency means that at $x=-2$, the curve is perfectly horizontal. Thus, we have two conditions locked in: $y(-2)=0$ and $y^{\prime }(-2)=0$. This is the secret key that unlocks the entire puzzle.

We are given the cubic function $y(x)=a{x}^3+b{x}^2+cx+5$. We have three unknowns: $a$, $b$, and $c$. To find them, we need three independent conditions.

The third condition comes from the y-intercept $Q$. At the y-axis, $x=0$, and the problem states the slope $y^{\prime }$ at this point is $3$. Let us differentiate our function:

Plugging in $x=0$, we get $y^{\prime }(0)=c=3$. Just like that, our first constant is revealed!

With $c=3$ in our pocket, we turn back to point $P(-2,0)$. Substituting $x=-2$ and $y=0$ into the original equation, we get:

This simplifies to the linear equation:

Next, we use the tangency condition $y^{\prime }(-2)=0$. Substituting $x=-2$ into our derivative $y^{\prime }(x)=3a{x}^2+2bx+3$, we get:

Adding these two linear equations, the $b$ terms vanish into thin air, leaving us with $4a+2=0$, so $a=-\frac{1}{2}$. Substituting this back, we find $b=-\frac{3}{4}$.

Now that we have our function $y(x)=-\frac{1}{2}{x}^3-\frac{3}{4}{x}^2+3x+5$, we are ready to find the local maximum. We set the derivative to zero:

Factoring out $-\frac{3}{2}$, we get $x^2+x-2=0$, which factors beautifully into $(x+2)(x-1)=0$. Our critical points are $x=-2$ and $x=1$.

We know $x=-2$ is our point of tangency (a local minimum in this specific cubic shape). That leaves $x=1$ as our candidate for the local maximum. A quick check with the second derivative $y^{\prime \prime }(x)=-3x-\frac{3}{2}$ confirms that $y^{\prime \prime }(1)=-4.5$, which is less than zero—a local maximum!

Finally, we calculate the value:

The journey is complete, and the final result is $\frac{27}{4}$.