Answer:
There is no such point where the given curve has a tangent line with slope 2.
Step-by-step explanation:
We have been given a curve [tex]y=4x^3+7x-5[/tex]. We are asked to show that the given curve has no tangent line with slope 2.
First of all, we will find the derivative of given curve as shown below:
[tex]y'=\frac{d}{dx}(4x^3)+\frac{d}{dx}(7x)-\frac{d}{dx}(5)[/tex]
[tex]y'=4\cdot 3x^{3-1}+7x^{1-1}-0[/tex]
[tex]y'=12x^{2}+7x^{0}[/tex]
[tex]y'=12x^{2}+7(1)[/tex]
[tex]y'=12x^{2}+7[/tex]
We know that derivative represents slope of tangent line, so we will equate derivative of the given curve with 2 and solve for the point (x), where the slope of tangent line will be equal to 2 as:
[tex]12x^2+7=2[/tex]
[tex]12x^2+7-7=2-7[/tex]
[tex]12x^2=-5[/tex]
[tex]x^2=-\frac{5}{12}[/tex]
We know that square of any real number could never be negative, therefore, there is no such point where the given curve has a tangent line with slope 2.
